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Need solution for RD Sharma maths class 12 chapter 21 Differential Equation exercise Fill in the blank question 19

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 Substituting the values


 x\frac{\mathrm{d} y}{\mathrm{d} x}+2y=x^{2}


On dividing the given differential equation by x, we get

\frac{\mathrm{d} y}{\mathrm{d} x}+2\frac{y}{x}=x

 So, the differential equation is a linear differential equation, and we can observe that the given differential equation is in the form

\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q


P=\frac{2}{x} \, and\, Q=x

Here we get the integrating factor as

I.F=e^{\int Pdx}

By substituting the values,

=e^{\int \frac{2}{x}dx}

We get,

\begin{aligned} &=e^{2log\, x} \\ &=e^{log\, x^{2}} \\ &=x^{2} \end{aligned}

Here, the general solution can be written as

\begin{aligned} &Y(I.F)=\int Q(I.F)dx+C \end{aligned}

Substituting the values we get

\begin{aligned} &y(x^{2})=\int x(x)^{2}dx+C \\ &x^{2}y=\frac{x^{4}}{4}+C \end{aligned}

Dividing the entire equation by x2

\begin{aligned} &y=\frac{x^{2}}{4}+\frac{C}{x^{2}} \\ \Rightarrow &y=\frac{1}{4}x^{2}+Cx^{-2} \end{aligned}

So, the answer is

\begin{aligned} &y=\frac{1}{4}x^{2}+Cx^{-2} \end{aligned}

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Gurleen Kaur

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